Question: Simplify the following expression: $y = \dfrac{2x^2+3x- 2}{x + 2}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(-2)} &=& -4 \\ {a} + {b} &=& &=& {3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-4$ and add them together. Remember, since $-4$ is negative, one of the factors must be negative. The factors that add up to ${3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${4}$ $ \begin{eqnarray} {ab} &=& ({-1})({4}) &=& -4 \\ {a} + {b} &=& {-1} + {4} &=& 3 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 {-1}x) + ({4}x {-2}) $ Factor out the common factors: $ x(2x - 1) + 2(2x - 1)$ Now factor out $(2x - 1)$ $ (2x - 1)(x + 2)$ The original expression can therefore be written: $ \dfrac{(2x - 1)(x + 2)}{x + 2}$ We are dividing by $x + 2$ , so $x + 2 \neq 0$ Therefore, $x \neq -2$ This leaves us with $2x - 1; x \neq -2$.